Navigate

The height at which the weight of a body becomes 1/16 th, its weight

The height at which the weight of a body becomes 1/16 th, its weight on the surface of earth (radius R), is

Options

(a) 5 R
(b) 15 R
(c) 3 R
(d) 4 R

Correct Answer:

3 R

Explanation:

Let at h height, the weight of a body becomes 1/16th of its weight on the surface.

Wh = 16 Ws g’ = GM/(Re+h)²

Similarly, g = GM/Re²

Now, g’/g = Re²/(Re+h)²

⇒ g’=g(1+h/R)⁻²

⇒ Wh = 16 Ws

⇒ g’ = g/16 g’

= 16g(1+h/R)⁻² 1/16

= (1+h/R)⁻²

⇒ 4 = 1+h/R

⇒ h = 3 R

AIPMT2012
admin:

Related Questions

  1. An electron moving in a uniform magnetic field of induction of intensity B,
  2. The energy of a hydrogen atom in the ground state is – 13.6 eV. The energy of a He⁺ ion
  3. A spherical liquid drop of radius R is divided into 8 equal droplets. If the surface
  4. If vectors A=cos wti + sin wtj and B= cos wt/2 i + sin wt/2 j are functions of time,
  5. A thin semicircular conducting ring (PQR) of radius r is falling with its plane